Problem: Find the slope and y-intercept of the line that is ${\text{parallel}}$ to $\enspace {y = -\dfrac{1}{3}x + 5}\enspace$ and passes through the point ${(-6, 4)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Explanation: Parallel lines have the same slope. The slope of the blue line is ${-\dfrac{1}{3}}$ , so the equation of our parallel line will be of the form $\enspace {y = -\dfrac{1}{3}x + b}\enspace$ We can plug our point, $(-6, 4)$ , into this equation to solve for ${b}$ , the y-intercept. $4 = {-\dfrac{1}{3}}(-6) + {b}$ $4 = 2 + {b}$ $4 - 2 = {b} = 2$ The equation of the parallel line is $\enspace {y = -\dfrac{1}{3}x + 2}\enspace$. ${m = -\dfrac{1}{3}, \enspace b = 2}$